Question

If 2.1 atms of PCl5(g) is introduced into an evacuated 3.00 L vessel and allowed to reach equilibrium at 250oC. PCl5(g) PCl3(g) + Cl2(g) Kp = 1.80 A. What is the value for Kc? B. What is the total pressure inside the vessel at equilibrium. Please show your work.

Answer 1

use:
Kp = Kc*(R*T)^(delta n)
PCl5(g) -->PCl3(g) + Cl2(g)
delta n = 1+1 -1 = 1
T=250 oC = (250+273) K =523 K
Kp = Kc*(R*T)^(delta n)
1.8 =Kc*(8.314*523)^1
Kc=4.14*10^-4

-------------------------------------------------
PCl5(g) -->PCl3(g) + Cl2(g)
2.1                  0                  0   (initial)
2.1-x               x                  x   (at eqiuilibrium)
Kp = x*x / (2.1-x)
1.8 = x*x / (2.1-x)
3.78 - 1.8x =x^2
x^2 + 1.8 x -3.78 =0
solving above quadratic equation, positive value of x is x=1.24 atm

Total pressure at equilibrium = 2.1-x + x + x =2.1+x =2.1 + 1.24 = 3.34 atm
Answer: 3.34 atm